Integrand size = 25, antiderivative size = 32 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=-\frac {b^4 F^a \Gamma \left (-4,-b (c+d x)^n \log (F)\right ) \log ^4(F)}{d n} \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2250} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=-\frac {b^4 F^a \log ^4(F) \Gamma \left (-4,-b (c+d x)^n \log (F)\right )}{d n} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = -\frac {b^4 F^a \Gamma \left (-4,-b (c+d x)^n \log (F)\right ) \log ^4(F)}{d n} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=-\frac {b^4 F^a \Gamma \left (-4,-b (c+d x)^n \log (F)\right ) \log ^4(F)}{d n} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(144\) vs. \(2(34)=68\).
Time = 0.64 (sec) , antiderivative size = 145, normalized size of antiderivative = 4.53
method | result | size |
risch | \(-\frac {F^{a} \left (\operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{n} \ln \left (F \right )\right ) \ln \left (F \right )^{4} b^{4} \left (d x +c \right )^{4 n}+F^{b \left (d x +c \right )^{n}} \ln \left (F \right )^{3} b^{3} \left (d x +c \right )^{3 n}+F^{b \left (d x +c \right )^{n}} \ln \left (F \right )^{2} b^{2} \left (d x +c \right )^{2 n}+2 F^{b \left (d x +c \right )^{n}} \ln \left (F \right ) b \left (d x +c \right )^{n}+6 F^{b \left (d x +c \right )^{n}}\right ) \left (d x +c \right )^{-4 n}}{24 n d}\) | \(145\) |
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Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (36) = 72\).
Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 3.72 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=\frac {{\left (d x + c\right )}^{4 \, n} F^{a} b^{4} {\rm Ei}\left ({\left (d x + c\right )}^{n} b \log \left (F\right )\right ) \log \left (F\right )^{4} - {\left ({\left (d x + c\right )}^{3 \, n} b^{3} \log \left (F\right )^{3} + {\left (d x + c\right )}^{2 \, n} b^{2} \log \left (F\right )^{2} + 2 \, {\left (d x + c\right )}^{n} b \log \left (F\right ) + 6\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{24 \, {\left (d x + c\right )}^{4 \, n} d n} \]
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\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=\int F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{- 4 n - 1}\, dx \]
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\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=\int { {\left (d x + c\right )}^{-4 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=\int { {\left (d x + c\right )}^{-4 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-4 n} \, dx=\int \frac {F^{a+b\,{\left (c+d\,x\right )}^n}}{{\left (c+d\,x\right )}^{4\,n+1}} \,d x \]
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